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Sin(3x)=3sinx ?

  1. Mar 30, 2004 #1
    This is a bizarre little proof that is probably wrong, but I want to know where
    [tex]\lim_{\theta\rightarrow0}\frac{\sin{3\theta}}{\theta} = 3[/tex]
    This is provable quite easily, I don't think I need to do that atm...
    well, we know that
    [tex]\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 1[/tex]
    So that would imply that
    [tex]3\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 3[/tex]
    thus implying that
    [tex]\lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta} = 3[/tex]
    By substitution
    [tex]\lim_{\theta \rightarrow 0} \frac{\sin{3\theta}}{\theta} = \lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta}[/tex]
    (iffy step)
    If you drop the limit on both sides...
    Which oh so quickly becomes
  2. jcsd
  3. Mar 30, 2004 #2

    Tom Mattson

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    Yes, that step is "iffy". With it, I can "prove" that sin(3θ) equals any function with the same limit as θ approaches zero. The step is invalid for the same reason that it is invalid to conclude that sin(3θ)=sin(θ) just because those two functions happen to be equal at θ=0.

    edit: fixed quote bracket
  4. Mar 30, 2004 #3
    Ok, cool, I can accept that. :)
  5. Mar 30, 2004 #4
    Sure, for certain values of theta.


    Edit: What Tom said.
  6. Mar 30, 2004 #5
    smartass :-p
  7. Mar 30, 2004 #6
    Hey now, I was just saying the same thing Tom was. ;)

    And getting to be a smartass is a privilege of being young!

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